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Problem 989

AIME late

Algebra Difficulty 6.0 Find the answer

B3. Solve the equation: $1+\log \left(2^{x}+1\right)=\log 2+\log \left(4^{x}+9\right)$ .

## 21st Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools National Competition, May 15, 2021

## Problems for 4th Year

Time for solving: 120 minutes. In section A, we will award three points for each correct answer, and deduct one point for each incorrect answer. Enter your answers for section A in the left table, leave the right table blank.
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Your answer

Official solution

B3. By applying the rule for the sum of logarithms, we transform the given equation into $\log 10\left(2^{x}+1\right)=$ $\log 2\left(4^{x}+9\right)$ . We antilogarithmize the equation to the form $10\left(2^{x}+1\right)=2\left(4^{x}+9\right)$ and, by introducing a new unknown $2^{x}=t$ , transform it into the equation $t^{2}-5 t+4=0$ . The solutions to the equation are $t_{1}=1$ and $t_{2}=4$ . We then calculate the values of the original unknowns of the equation: $x_{1}=0$ and $x_{2}=2$ .

# dUFA

21st competition in mathematics for students of secondary technical and vocational schools National competition, May 15, 2021 Solutions to problems for 4th year

$\begin{array}{|c|c|c|} \hline \mathrm{A} 1 & \mathrm{~A} 2 & \mathrm{~A} 3 \\ \hline \mathrm{E} & \mathrm{C} & \mathrm{C} \\ \hline \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.